In this lesson, we will deal with quadratic equations, which are polynomial equations of the second degree. What this means is that when the equation is expanded out with no fractional terms, there exists at least one term in which the unknown quantity is raised to the second power.

The general form of a quadratic equation is ax^2 + bx + c. The quadratic formula can be used to solve equations in the standard form. The quadratic formula is usually written as below:

In this lesson, however, we will not deal with quadratic equations that are written in the standard form. It may take quite a bit of effort in cross-multiplication, collection of like terms, etc., to get these equations into the standard form. The solution of these equations using the quadratic formula, therefore, is quite cumbersome, not to mention, error-prone.

Rather than go through all that labor, we will identify these special types of equations and apply certain simple procedures to solve them quickly and easily.

The first special type of quadratic equations we will consider are like the one below:

x + 1/x = 10/3

In the traditional method, we would go through the process of getting the equation into the standard form using the steps below:

We would then solve it using the quadratic formula by setting a = 3, b = -10 and c = 3. This would lead to the solutions x = 3 and x = 1/3.

However, we don't have to go through all that trouble to solve this kind of equation. All we have to do is observe that 10/3 = 3 + 1/3. Thus, we would immediately have figured out that we can rewrite the given equation as below:

We are not restricted to x + 1/x on the left-hand side either. Consider the equation:

(2x + 3) + 1/(2x + 3) = 50/7

Since the right-hand side of the equation can be expanded to 7 + 1/7, by the symmetry of the equation, we can equate 2x + 3 to either 7 or 1/7 (or equivalently, by equating (2x + 3) and 1/(2x + 3) to 7). We then get the solutions x = 2 and x = -10/7 to the given equation.

Similarly, consider the equation:

5x/(2x + 3) + (2x + 3)/5x = 26/5

The right-hand side of the equation can be expanded to 5 + 1/5. Thus, we can use the symmetry of the resulting equation to derive the following linear equations:

5x/(2x + 3) = 55x/(2x + 3) = 1/5 or alternatively, (2x + 3)/5x = 5

These two equations can then be solved to give us x = -3 or x = 23/3.

Now, consider the equation:

(x + 3)/(3x + 5) + (3x + 5)/(x + 3) = 17/4

The right-hand side of the equation can be expanded to 4 + 1/4. This then lets us solve the equation by deriving the linear equations below:

We may be tempted to conclude from the symmetry of the equation on both sides of the equal-to sign that x = 2 or x = 1/2. That would be wrong. In equations such as the above where the terms are connected by "-" signs instead of "+" signs, the solutions are x = 2 and x = -1/2. Only with x = -1/2 is it possible to get -1/x = 2, and therefore x - 1/x = 2 - 1/2. This is important to remember.

Many of these transformations are difficult to peal back on sight to reveal the true nature of the equation. But it may be worthwhile to examine the equation and try a few transformations to see if any simplifications are possible before giving up and solving it using the traditional method.

Sometimes, the right-hand side is not as easily decomposed to a pair of reciprocals as in the above examples. Consider the equation below:

x + 1/x = 25/12

The right-hand side may look as if it can not be converted into a pair of reciprocals since at first glance, it breaks apart as 2 and 1/12. But a closer look will reveal that 25/12 = 3/4 + 4/3. Thus, we can actually rewrite the equation above as:

x + 1/x = 3/4 + 4/3

This then leads immediately to the solutions x = 3/4 and x = 4/3 by the symmetry of the equation on both sides of the equal-to sign.

This same technique can be extended to equations where the left-hand side is a sum or difference of any two reciprocal quantities, not just x and 1/x. I will leave those extensions to the reader in the interest of keeping this lesson from growing any longer than it already has become!

Assuming that the left-hand side is a pair of reciprocals, connected by either "+" or "-", how do we verify whether the right-hand side can be expressed as a pair of reciprocals with the same sign between them? We will explore this question in greater detail in the next lesson.

In the meantime, I hope you have found this lesson useful and interesting. I also hope you will apply the techniques explained in this lesson on real problems so that you become familiar not only with the technique itself (which is actually quite trivial), but also with the fractions that result either from the addition of numbers with their reciprocals, or the differences between numbers and their reciprocals. That will enable one to apply this technique where appropriate, on sight and mentally, to solve the types of quadratic equations we have dealt with in this lesson. Good luck, and happy computing!

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